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Help with finding normal modes of a bar swinging on a string

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A bar with mass m and length l is attached at one end to a string (also length l) and is swinging back and forth. Find the normal modes of oscillation.


    2. Relevant equations
    L=T-U, and the Lagrange-euler equation, I=(1/12)ml^2


    3. The attempt at a solution
    So my idea is this. Use the two angles [itex]\theta[/itex] (angle of string from the vertical) and [itex]\phi[/itex] (angle of the bar from the vertical) as the generalized coordinates and set up a Lagrangian to get two (probably coupled) differential equations. THen I'm guessing two different eigenfrequencies will pop out when I solve them.

    My problem is that I'm not sure what the (translational) kinetic energy is. (The rotational would just be I*ω^2 and this ω^2 would, in the end, give me the two frequencies [?]) I've tried finding [itex]\dot{x}[/itex] and [itex]dot{y}[/itex] using sines and cosines of the angles, but when I do (and I make appropriate approximations) I'm left with no [itex]dot{\theta}[/itex] or [itex]dot{\phi}[/itex]terms which doesn't seem right.
    But I'm also not sure if it's as simple as [itex]T=ml^2(dot{\theta}+dot{phi}) [/itex].

    Thanks
     
    Last edited: Apr 9, 2013
  2. jcsd
  3. Apr 9, 2013 #2

    vela

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    The easiest thing to do is simply write down the x and y coordinates of the center of mass of the rod in terms of the two angles and then calculate ##\frac{1}{2}(\dot{x}^2+\dot{y}^2)##.
     
  4. Apr 9, 2013 #3
    Thanks. A new question though:

    What is ω (from the rotational KE) in terms of the two angles? Usually [itex]\omega = dot{\theta} [/itex]. But that's only for one angle. Because if ω isn't in terms of the angle(s) then the rotational KE drops out when you take the derivatives for L which doesn't seem right.

    Thanks again.
     
  5. Apr 9, 2013 #4

    vela

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    From the way you defined the angles, you should have ##\omega=\dot{\phi}##, right? Changing that angle corresponds to the bar rotating whereas changing ##\theta## only causes translation of the bar.
     
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